Resources
1. Notes
2. NCERT Solutions
3. Worksheets
4. PYQs
5. Chapter Test
Resources
Home > Class 10 > Chapter 5: Arithmetic Progressions

Chapter 5: Arithmetic Progressions

Overview

This page provides comprehensive Chapter 5: Arithmetic Progressions – Board Exam Notes aligned with the latest CBSE 2025–26 syllabus. Covers nth term of an AP, sum of first n terms, common difference, and real-life situational problems.

nth Term • Sum of n Terms • Common Difference • Word Problems

Exam Weightage & Blueprint

Total: 6-7 Marks

This chapter falls under Unit II: Algebra (20 marks total). As per the latest syllabus, focus is on: derivation of the nth term and sum of first n terms of an AP and their application in solving daily life problems.

Question Type Marks Frequency Focus Topic
MCQ 1 High Finding 'd', 'a', or nth term
Short Answer 2 or 3 Medium Sum of AP ($S_n$), finding $n$
Case Study 4 High Real-life Applications (Ladders, Savings)

⏰ Last 24-Hour Checklist

  • Definition: Fixed difference between terms.
  • Common Difference (d): $a_2 - a_1$.
  • General Form: $a, a+d, a+2d...$
  • nth Term ($a_n$): $a + (n-1)d$.
  • Sum ($S_n$): $\frac{n}{2}[2a+(n-1)d]$.
  • nth Term from Sum: $a_n = S_n - S_{n-1}$.

📖 Concepts & Definitions

Arithmetic Progression (AP): A list of numbers in which each term is obtained by adding a fixed number to the preceding term except the first term.

Key Components

First Term (a)

The starting number of the sequence.

Difference (d)

$d = a_{k+1} - a_k$. Can be $+$, $-$, or $0$.

General Term ($a_n$)

Value of the term at position $n$.

Check for AP: To check if a list is an AP, calculate differences ($a_2-a_1$, $a_3-a_2$). If they are EQUAL, it's an AP.

🧮 Important Formulas

1. General Term

$$ a_n = a + (n - 1)d $$

2. nth Term from the END (Exam Secret)

If $l$ is the last term, use this direct formula instead of reversing the AP:

$$ \text{nth term from end} = l - (n - 1)d $$

3. Sum of First n Terms ($S_n$)

$$ S_n = \frac{n}{2} [ 2a + (n - 1)d ] $$
OR (if last term $l$ is known)
$$ S_n = \frac{n}{2} ( a + l ) $$

4. Sum of First n Positive Integers

The sum of the first $n$ natural numbers ($1 + 2 + 3 + ... + n$) is a special AP sum:

$$ S_n = \frac{n(n + 1)}{2} $$

5. Arithmetic Mean (AM)

If three numbers $a, b, c$ are in AP, then the middle term $b$ is the arithmetic mean of the other two:

$$ b = \frac{a + c}{2} \Rightarrow 2b = a + c $$

6. Finding Term from Sum (High-Yield HOTS)

If $S_n$ is given (e.g., $S_n = 4n - n^2$) and you need to find the AP or $a_n$:

$$ a_n = S_n - S_{n-1} \quad (\text{for } n > 1), \qquad a_1 = S_1 $$ $$ d = a_2 - a_1 = (S_2 - S_1) - S_1 = S_2 - 2S_1 $$

7. Standard Selection of Terms in AP

When the sum of consecutive terms in an AP is given, choose terms symmetrically so that $d$ cancels out when adding:

Number of Terms Terms to Choose Common Difference
3 terms $a - d, \; a, \; a + d$ $d$ (Sum $= 3a$)
4 terms $a - 3d, \; a - d, \; a + d, \; a + 3d$ $2d$ (Sum $= 4a$)
5 terms $a - 2d, \; a - d, \; a, \; a + d, \; a + 2d$ $d$ (Sum $= 5a$)
💡 First Negative Term Shortcut:
To find the first negative term of a decreasing AP (e.g., $20, 19\frac{1}{4}, 18\frac{1}{2}, \dots$):
Set $a_n < 0 \implies a + (n-1)d < 0$. Solve the inequality for $n$ and take the smallest integer greater than $n$.
Solve Exercise 5.3 (Sum of first n terms) Solve Exercise 5.4 (Optional & Advanced)

Solved Examples (Board Marking Scheme)

Q1. Find the 11th term from the last term of AP: 10, 7, 4, ..., -62. (2 Marks)

Step 1: Identify Parameters 0.5 Mark

$a=10$, $d = 7-10 = -3$, Last term $l = -62$, $n=11$.

Step 2: Apply Formula 1 Mark

nth term from end $= l - (n-1)d$

$= -62 - (11-1)(-3)$

$= -62 - (10)(-3)$

Step 3: Calculate 0.5 Mark

$= -62 + 30 = -32$.

Q2. If the sum of first n terms is $S_n = 4n - n^2$, find the 10th term. (3 Marks)

Step 1: Find S10 and S9 1.5 Marks

$S_{10} = 4(10) - (10)^2 = 40 - 100 = -60$.

$S_9 = 4(9) - (9)^2 = 36 - 81 = -45$.

Step 2: Apply Formula 1 Mark

$a_n = S_n - S_{n-1}$

$a_{10} = S_{10} - S_9$

$a_{10} = -60 - (-45)$

Step 3: Answer 0.5 Mark

$a_{10} = -60 + 45 = -15$.

Q3. Which term of the AP: 21, 18, 15, ... is -81? (3 Marks)

Step 1: Identify Parameters 0.5 Mark

$a = 21$, $d = 18 - 21 = -3$, $a_n = -81$.

Step 2: Apply Formula 1 Mark

$-81 = 21 + (n-1)(-3)$

$-102 = -3(n-1)$

Step 3: Solve 1.5 Marks

$34 = n - 1 \Rightarrow n = 35$.

Therefore, the 35th term is -81.

Previous Year Questions (PYQs)

2023 (1 Mark): Find the 10th term of AP: 2, 7, 12...
Ans: $a=2, d=5, n=10$. $a_{10} = 2 + 9(5) = 47$.
2020 (3 Marks): How many two-digit numbers are divisible by 3?
Ans: Sequence: 12, 15, ..., 99. Here $a=12, d=3, a_n=99$. Solving $99=12+(n-1)3$ gives $n=30$.
2019 (4 Marks): The sum of 4th and 8th terms of an AP is 24 and sum of 6th and 10th terms is 44. Find the AP.
Ans: Eq1: $a+3d + a+7d = 24 \Rightarrow 2a+10d=24$. Eq2: $2a+14d=44$. Solving gives $d=5, a=-13$. AP: -13, -8, -3...

Exam Strategy & Mistake Bank

⚠️ Mistake Bank

Common Difference Sign: If AP is decreasing (10, 8, 6...), $d$ must be negative ($-2$). Students often write $+2$.
n vs nth term: Confusing "number of terms" ($n$) with "value of term" ($a_n$).
Calculation: In $S_n$ formula, forgetting to multiply $n/2$ with the bracket content.

💡 Scoring Tips

Write Formula: Always write the generic formula ($a_n = ...$) before substituting values. It carries marks.
Case Studies: For word problems (flower beds, savings), list the sequence first (e.g., 20, 25, 30...) then apply AP formulas.
n must be integer: If you calculate $n$ as a fraction (e.g., 15.5), recheck your work. Number of terms cannot be decimal.

Self-Assessment Mock Test (10 Marks)

Q1 (1M): Write the common difference of AP: $3, 1, -1, -3\dots$

Q2 (2M): Find the 31st term of an AP whose 11th term is 38 and 16th term is 73.

Q3 (3M): How many terms of the AP: $9, 17, 25\dots$ must be taken to give a sum of 636? Explain the double answer if any.

Q4 (4M): If the sum of first $n$ terms is $S_n = 4n - n^2$, find the first term, the 2nd term, the common difference, and the $n$-th term.

📝 More Solved Board Questions

Q4. The 4th term of an AP is 18. The difference of the 9th term and the 15th term is 30. Find the AP. 3 Marks

Sol. Given: $a_4 = 18 \Rightarrow a + 3d = 18$   ...(1)

Also, $a_{15} - a_9 = 30$

$(a + 14d) - (a + 8d) = 30 \Rightarrow 6d = 30 \Rightarrow d = 5$

Put $d=5$ in (1): $a + 3(5) = 18 \Rightarrow a = 3$

AP: 3, 8, 13, 18, ...

Q5. Find the sum of all two-digit odd positive numbers. 3 Marks

Sol. Two-digit odd numbers: 11, 13, 15, ..., 99

Here $a = 11$, $d = 2$, $a_n = 99$.

Find $n$: $99 = 11 + (n-1)2 \Rightarrow 88 = 2(n-1) \Rightarrow n-1 = 44 \Rightarrow n = 45$.

Sum: $S_{45} = \frac{45}{2}(11 + 99) = \frac{45}{2}(110) = 45 \times 55 = 2475$.

Answer: 2475

Q6. Which term of the AP: $121, 117, 113\dots$ is its first negative term? 3 Marks (Standard Board)

Sol. Here $a = 121, d = 117 - 121 = -4$.

Let $n$-th term be negative: $a_n < 0$

$a + (n - 1)d < 0 \implies 121 + (n - 1)(-4) < 0$

$121 - 4n + 4 < 0 \implies 125 < 4n \implies n > \frac{125}{4} = 31.25$

Since $n$ must be a natural number, the least value of $n$ is $32$.

Therefore, the 32nd term is the first negative term.

Q7. The sum of three numbers in AP is 21 and their product is 231. Find the numbers. 4 Marks (HOTS)

Sol. Let the three numbers be $(a - d), a, (a + d)$.

Sum $= (a - d) + a + (a + d) = 3a = 21 \implies \mathbf{a = 7}$.

Product $= (7 - d)(7)(7 + d) = 231 \implies 7(49 - d^2) = 231$

$49 - d^2 = 33 \implies d^2 = 49 - 33 = 16 \implies \mathbf{d = \pm 4}$.

When $d = 4$: numbers are $7 - 4, 7, 7 + 4 \implies \mathbf{3, 7, 11}$.

When $d = -4$: numbers are $7 - (-4), 7, 7 - 4 \implies \mathbf{11, 7, 3}$.

Hence, the numbers are 3, 7, 11.

🎯 Board Pattern (2018–2025): Word problems regarding savings (e.g., Subba Rao saving ₹5 in the first week) or building structures (e.g., number of logs in rows) are frequently asked as Case Study questions. Always identify $a$ and $d$ from the context first.

📋 Board Revision Checklist

  • ✅ Identify if a sequence is an AP by checking $a_2 - a_1 = a_3 - a_2 = d$
  • ✅ Formula for nth term: $a_n = a + (n-1)d$
  • ✅ Formula for sum of first n terms: $S_n = \frac{n}{2}[2a + (n-1)d]$
  • ✅ Direct sum formula when last term is known: $S_n = \frac{n}{2}(a + l)$
  • ✅ Direct formula for nth term from end: $l - (n-1)d$
  • ✅ First negative term: solve $a_n < 0 \implies n > \text{fraction}$, pick the next integer
  • ✅ Natural number sum: $1 + 2 + \dots + n = \frac{n(n+1)}{2}$
  • ✅ Remember: $n$ must always be a positive integer ($n \in \mathbb{N}$)
  • ✅ Relationship between $a_n$ and $S_n$: $a_n = S_n - S_{n-1}$ and $a_1 = S_1$
  • ✅ Three terms in AP with sum given: use $(a-d), a, (a+d)$
  • ✅ Arithmetic Mean: if $a, b, c$ are in AP, then $2b = a + c$
💡 Exam Tip:
In word problems, if the question asks "In which year..." or "How many terms...", you are solving for $n$. If it asks "What is the total amount...", you are solving for $S_n$.

Concept Mastery Quiz 🎯

Test your readiness for the board exam.

1. The common difference of the AP: $5, 2, -1, -4...$ is:

2. If the first term is 10 and common difference is 10, the 4th term is:

3. The sum of first 100 natural numbers is:

4. If $k, 2k-1, 2k+1$ are in AP, the value of $k$ is:

5. In an AP, if $d = -4, n = 7, a_n = 4$, then $a$ is:

PRACTICE THIS CHAPTER
← Previous Chapter 4 Next Chapter 6 →